This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 , c 2 , and c 3 such that c 1 u + c 2 v + c 3 w = (2, 0, 4) u = (3, 1, 2), v = (4, 0, 8), and w = (6, 1, 4) This means that we have 3 equations in 3 unknowns: Equation 1: 3c 1 + 4c 2 + 6c 3 = 2 Equation 2:c 1 c 3 = 0 Equation 3: 2c 1 – 8c 2 – 4c 3 = 4 From Equation 2: c 1 = c 3 , substituting we get: Equation 1: 3c 1 + 4 c 2 = 2 Equation 3: 2c 1 – 8c 2 = 4, or  c 1 – 4c 2 = 2. Adding the two equations, we get: 2c 1 = 4, c 1 = c 3 = 2. Subs. In Equation 1: 6 + 4c 2 = 2, 4c 2 = 4, c 2 = 1. Answer: c 1 = 2, c 2 = 1, and c 3 = 2...
View
Full
Document
This note was uploaded on 10/01/2009 for the course ENGR 1100 taught by Professor Anderson during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Anderson

Click to edit the document details